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\usepackage{ctex} % 中文支持
\usepackage{amsmath, amsthm, amssymb, bm} % 数学公式与符号
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{booktabs} % 用于高质量表格
\usepackage{tikz} % 可选：用于绘图
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\usepackage{pythonhighlight}

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\usecolortheme{default} % 可选：default, seahorse, beaver, dolphin 等

% 自定义定理样式
\setbeamertemplate{theorems}[numbered]
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\newtheorem{mytheorem}{定理}
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\newtheorem{mycorollary}{推论}
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% 信息设置
\title[幂级数解法]{《常微分方程》第七章：幂级数解法}
\author[]{LQW}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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\begin{document}

% 封面页
\begin{frame}
  \titlepage
\end{frame}

% 目录页
%\begin{frame}{目录}
%  \tableofcontents
%\end{frame}

%\maketitle

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\begin{frame}{目录}

\begin{enumerate}
\item[7.1.]   柯西定理 
\item[7.2.]   幂级数解法
\item[7.3.]   勒让德多项式
\item[7.4.]   广义幂级数解法
\item[7.5.]   贝塞尔函数

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.1. 二元解析函数}

\begin{itemize}

\item  {\color{red}问：称二元函数 $f(x,y)$ 在平面区域 $G$ 内解析，是指什么？}

\item  答：是指对区域 $G$ 内任意一点 $(x_0,y_0)$, 存在正数 $a$ 和 $b$, 使得在矩形区域 $$|x-x_0|< a, \,\, |y-y_0|< b$$
中，这个函数可以写成收敛的幂级数的形式：$$f(x,y)=\sum\limits_{i,j=0}^{\infty} a_{ij}(x-x_0)^i(y-y_0)^j.$$


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.2. 柯西定理 }

\begin{itemize}

\item  {\color{red}柯西定理：设 $f(x,y)$ 在平面区域 $G$ 内解析，设 $(x_0,y_0)\in G$, 则初值问题 $$\frac{dy}{dx}=f(x,y),\,\, y(x_0)=y_0$$
在 $x_0$ 的某个领域内的解函数也是解析的，即可以写成收敛幂级数形式：$$y=\sum\limits_{n=0}^{\infty} C_n(x-x_0)^n.$$
}

%\item  答：


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.3. 优级数 }

\begin{itemize}

\item  {\color{red}问：什么是优级数？}

\item  答：如果对任意 $i,j$ 都有 $|a_{ij}|\le A_{ij}$, 那么称幂级数 
\begin{eqnarray*}
\sum\limits_{i,j=0}^{\infty} A_{ij}(x-x_0)^i(y-y_0)^j
\end{eqnarray*}
是幂级数
\begin{eqnarray*}
\sum\limits_{i,j=0}^{\infty} a_{ij}(x-x_0)^i(y-y_0)^j
\end{eqnarray*}
的一个优级数。


\end{itemize}

\end{frame}

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\begin{frame}{7.1.4. 优函数}

\begin{itemize}

\item  {\color{red}问：称 $F(x,y)$ 是 $f(x,y)$ 的一个优函数，是指什么？}

\item  答：
\begin{enumerate} 
\item  在区域 $R$ 内有收敛的幂级数展开
\begin{eqnarray*}
F(x,y) &=& \sum\limits_{i,j=0}^{\infty} A_{ij}(x-x_0)^i(y-y_0)^j, \\ 
f(x,y) &=& \sum\limits_{i,j=0}^{\infty} a_{ij}(x-x_0)^i(y-y_0)^j,  
\end{eqnarray*}

\item  对任意 $i,j$ 都有 $|a_{ij}|\le A_{ij}$. 
\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.5. 收敛幂级数均有“标准”优函数}

\begin{itemize}

\item  {\color{red}引理7.1：设函数 $f(x,y)$ 在矩形区域  $R: |x|< \alpha, \,\, |y|< \beta$ 中可以展开成一个收敛的幂级数，则存在正数 $M$ 与 $a<\alpha$ 与 $b<\beta$, 使得函数 $$F(x,y) = \frac{M}{(1-\frac{x}{a})(1-\frac{y}{b})}$$ 在矩形区域 $R_0 = \{(x,y): |x|< a, \,\, |y|< b\}$ 中是 $f(x,y)$ 的一个优函数。
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.6. 证明：收敛幂级数均有“标准”优函数}

\begin{enumerate}

\item  设展开的幂级数为   
$%\begin{eqnarray*}
f(x,y) = \sum\limits_{i,j=0}^{\infty} a_{ij}x^iy^j. 
$%\end{eqnarray*}

\item  因为这个幂级数在矩形区域 $R$ 中收敛，所以对任意 $a<\alpha$, $b<\beta$, 级数    
$%\begin{eqnarray*}
\sum\limits_{i,j=0}^{\infty} a_{ij}a^ib^j  
$%\end{eqnarray*}
是收敛的。

\item  因为收敛幂级数的通项是有界的，所以存在常数 $M>0$ 使得 $ |a_{ij}a^ib^j| \le M $ 对任意 $i,j$ 都成立。

\item  因为 $|a_{ij}| \le \frac{M}{a^ib^j}$, 所以 
$%\begin{eqnarray*}
F(x,y) = \sum\limits_{i,j=0}^{\infty} \frac{M}{a^ib^j} x^iy^j 
$%\end{eqnarray*}
是 $\sum\limits_{i,j=0}^{\infty} a_{ij}x^iy^j$ 的一个优级数。

\item  因为这个幂级数在区域 $R_0$ 收敛，所以 $F(x,y)$ 是 $f(x,y)$ 的一个优函数。

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.7. }

\begin{itemize}

\item  {\color{red} 引理7.2： 考虑定义在矩形区域 $R_0: |x|< a, \,\, |y|< b$ 中的二元函数 $$F(x,y) = \frac{M}{(1-\frac{x}{a})(1-\frac{y}{b})}.$$  则初值问题 $$\frac{dy}{dx} = F(x,y),\,\, y(0)=0$$ 在区间 $|x|<\rho$ 内存在一个解析解，其中 $$\rho=a(1-e^{- \frac{b}{2aM} } ).$$
}

\end{itemize}

\end{frame}

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\begin{frame}{7.1.8. 证明引理7.2 }

\begin{enumerate}

\item  为简化计算，考虑 $M=1, a=1, b=1$. 求解微分方程 $$\frac{dy}{dx} = \frac{1}{(1-x)(1-y)}.$$
\item  分离变量可得 $$ (1-y)dy = \frac{dx}{1-x}. $$
\item  两边积分，根据初值条件 $y(0)=0$, 可得 $$y-\frac{1}{2}y^2 = \ln(1-x). $$
\item  当 $|x| < 1- \frac{1}{\sqrt{e}} $ 时，上式定义了一个收敛的幂级数 $y(x)$. 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.9. 柯西定理}

\begin{itemize}

\item  {\color{red} 定理7.1. 如果函数 $f(x,y)$ 在矩形区域 $R: |x-x_0|<\alpha, |y-y_0|<\beta$ 上可以展开成收敛幂级数，则存在正数 $M,a,b$ 使得初值问题
$$\frac{dy}{dx} = f(x,y),\,\, y(x_0)=y_0$$
在领域 $|x-x_0|<\rho$ 内存在唯一的解析解，其中 $$ \rho=a[1-\exp(-b/2aM)]. $$
}


\end{itemize}

\end{frame}

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\begin{frame}{7.1.10. 证明柯西定理 }

\begin{enumerate}

\item  不妨设 $x_0=0, y_0=0$. 考虑初值问题 $$\frac{dy}{dx} = \sum\limits_{i,j=0}^{\infty} a_{ij}x^iy^j,\,\, y(0)=0. $$

\item  根据初值条件，设形式幂级数解如下，其中系数 $C_k$ 待定， $$ y = \sum\limits_{k=1}^{\infty} C_kx^k. $$

\item  代入微分方程，左边逐项求导，右边代入 $y$ 的形式幂级数，可得 $$ \sum\limits_{k=1}^{\infty} kC_kx^{k-1} = \sum\limits_{i,j=0}^{\infty} a_{ij}x^i 
%\left(\sum\limits_{k=1}^{\infty} C_kx^k \right)^j. 
\left( C_1x + C_2x^2 + C_3x^3 + \cdots + C_kx^k + \cdots \right)^j. 
$$

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.11. 证明柯西定理（续）}

\begin{enumerate}\setcounter{enumi}{3}%\itemsep0.5em

\item  比较等式两边的常数项，可得 $C_1=a_{00}$. 

\item  比较等式两边的1次项，可得 $2C_2 = a_{10} + a_{01}C_1$. 

\item  比较等式两边的2次项，可得 $3C_3 = a_{20} + a_{11}C_1 + a_{02}C_1^2 + a_{01}C_2$. 

\item  一般地，可以递推地将 $C_k$ 写成 $a_{ij}$ 的多项式。

\item  不妨设 $\sum\limits_{i,j=0}^{\infty} |a_{ij}|$ 收敛。因此通项系数有界，即存在常数 $M$ 使得 $|a_{ij}|\le M$. 

\item  考虑另一个初值问题，方程右端是原方程右端的一个优级数，
$$\frac{dy}{dx} = \sum\limits_{i,j=0}^{\infty} \hat{a}_{ij} x^iy^j= \frac{M}{(1-x)(1-y)},\,\, y(0)=0. $$

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.12. 证明柯西定理（续）}

\begin{enumerate}\setcounter{enumi}{9}%\itemsep0.5em

\item  同样有形式幂级数解，$$ y = \sum\limits_{k=1}^{\infty} \hat{C}_kx^k. $$

\item  根据 $C_k$ 是 $a_{ij}$ 的正系数多项式， $\hat{C}_k$ 是 $\hat{a}_{ij}$ 的正系数多项式，以及 $a_{ij}\le \hat{a}_{ij}$, 可知两个形式幂级数解的系数也成立同样的大小关系，即有 $$ C_k\le \hat{C}_k. $$ 

\item  因为 $\sum\limits_{k=1}^{\infty} \hat{C}_kx^k$ 在给定区间内是收敛的，所以 $\sum\limits_{k=1}^{\infty} C_kx^k$
在同样区间内也是收敛的。 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.13. }

\begin{itemize}

\item  {\color{red} 问题：举例说明，如果 $f(x,y)$ 在 $(x_0,y_0)$ 附近不是一个解析函数，那么微分方程初值问题 
$$\frac{dy}{dx}=f(x,y),\,\, y(x_0)=y_0$$
在 $x_0$ 附近不一定有形式幂级数解。
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.14. 附注1. }

\begin{itemize}

\item  {\color{red} 问题：验证下述微分方程初值问题在 $x_0=0$ 附近没有形式幂级数解：
$$x\frac{dy}{dx} = y-x,\,\, y(0)=0.$$
}

\vspace{-0.3cm}

\item  解答：
\begin{enumerate}
\item  设形式幂级数解为 $ y = C_1x + C_2x^2 + C_3x^3 + \cdots + C_nx^n + \cdots. $

\item  代入微分方程可得 
\begin{eqnarray*}
&& x(C_1 + 2C_2x + 3C_3x^2 + \cdots + nC_nx^{n-1} + \cdots) \\ 
&=& (C_1-1)x + C_2x^2 + C_3x^3 + \cdots + C_nx^n + \cdots. 
\end{eqnarray*}

\item  比较同次项系数可知无解。
\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.15. 附注2. }

\begin{itemize}

\item  {\color{red} 问题：验证下述初值问题在 $x_0=0$ 附近有形式幂级数解，但不收敛：
$$x^2\frac{dy}{dx}=y-x,\,\, y(0)=0.$$
}

\vspace{-0.3cm}

\item  解答：
\begin{enumerate}
\item  设形式幂级数解为 $ y = C_1x + C_2x^2 + C_3x^3 + C_4x^4 + \cdots + C_nx^n + \cdots. $

\item  代入微分方程可得 
\begin{eqnarray*}
&& x^2(C_1 + 2C_2x + 3C_3x^2 + C_4x^4 +\cdots + nC_nx^{n-1} + \cdots) \\ 
&=& (C_1-1)x + C_2x^2 + C_3x^3 + 4C_4x^4 +\cdots + C_nx^n + \cdots. 
\end{eqnarray*}

\item  比较同次项系数可知 $ y = x + x^2 + 2x^3 + 3!x^4 + \cdots + (n-1)!x^n + \cdots. $

\item  这个级数只在 $x=0$ 时收敛。%的收敛半径是零。


\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.1.16. Augustin Louis Cauchy (1789 - 1857) }


\begin{center}
\includegraphics [height=0.66\textheight, width=0.55\textwidth]{augustin-cauchy-stamp.jpg}
\end{center}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.1. 微分方程的常点与奇点}

\begin{itemize}\itemsep0.5em

\item  {\color{red} 问题：设函数 $A(x),B(x),C(x)$ 在 $x=x_0$ 附近解析。考虑微分方程 
$$ A(x)y''+B(x)y'+Q(x)y=0. $$
什么时候称 $x_0$ 是它的常点？什么时候称 $x_0$ 是它的奇点？}

\item  解答：
\begin{enumerate}\itemsep0.5em

\item  如果在 $x_0$ 附近，函数 $\frac{B(x)}{A(x)}$ 和 $\frac{C(x)}{A(x)}$ 都是解析的，那么称 $x_0$ 是这个微分方程的常点。否则就称 $x_0$ 是这个微分方程的奇点。

\item  例子：微分方程 $x^2y''+xy'+y=0$ 有一个奇点 $x_0=0$, 其余 $x_0$ 都是常点。

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.2. 幂级数解法 }

\begin{itemize}\itemsep0.5em

\item  {\color{red} 定理7.2：设二阶齐次线性微分方程 $$y''+p(x)y'+q(x)y=0$$ 的系数 $p(x)$ 和 $q(x)$ 在区间 $|x-x_0|<r$ 内可以展开成收敛的幂级数，那么通解在这个区间内也可以写成收敛幂级数 $$y=C_0+C_1(x-x_0) + C_2(x-x_0)^2 +\cdots+ C_n(x-x_0)^n+\cdots $$
的形式，其中 $C_0$ 和 $C_1$ 是两个任意常数。}

\item  证明：应用微分方程组的柯西定理。记 $z=y'$, 可得微分方程标准形式
\begin{eqnarray*}
\frac{d}{dx}\begin{bmatrix} y \\ z \end{bmatrix} 
= \begin{bmatrix} 0 & 1 \\ -q(x) & -p(x) \end{bmatrix} 
\begin{bmatrix} y \\ z \end{bmatrix}. 
\end{eqnarray*}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.3. }

\begin{itemize}\itemsep0.5em

\item  {\color{red} 例子7.2.1：用幂级数方法求解 Airy 方程 $y''=xy, \,\, -\infty<x<\infty$. } 

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  设幂级数解为 $y=a_0+a_1x+a_2x^2+a_3x^3 +\cdots + a_nx^n +\cdots$. 

\item  代入方程，可得 
\begin{eqnarray*}
&& (a_0+a_1x+a_2x^2+a_3x^3+\cdots + a_nx^n + \cdots )'' \\ 
&=&  x(a_0+a_1x+a_2x^2+a_3x^3+\cdots + a_nx^n + \cdots ). 
\end{eqnarray*}

\item  比较同次项的系数，可得系数的递推关系式 $$a_{n+2} = \frac{a_{n-1}}{(n+2)(n+1)}. $$

\item  得到两个线性无关的幂级数解。两个幂级数解的前四项为：

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.4. }

\begin{itemize}\itemsep0.5em

\item  {\color{red} 例子7.2.2：求 Airy 方程 $y''=xy$ 在 $x=1$ 附近的幂级数解。 } 

\item  解答：
\begin{enumerate}\itemsep0.3em

\item  设 $t=x-1$, 原方程化为 $y''=ty+y$. 

\item  设幂级数解为 $y=a_0+a_1t+a_2t^2+a_3x^3 +\cdots + a_nt^n +\cdots$. 

\item  代入方程，可得 
\begin{eqnarray*}
&& (a_0+a_1t+a_2t^2+a_3x^3+\cdots + a_nt^n + \cdots )'' \\ 
&=&  (t+1)(a_0+a_1t+a_2t^2+a_3x^3+\cdots + a_nt^n + \cdots ). 
\end{eqnarray*}

\item  比较同次项的系数，可得系数的递推关系式 $$(n+2)(n+1)a_{n+2} = a_n + a_{n-1}. $$

\item  得到两个线性无关的幂级数解。两个幂级数解的前四项为：

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.5. George Biddell Airy FRS (1801 - 1892) }

\begin{center}
\begin{figure}
\includegraphics [height=0.65\textheight, width=0.3\textwidth]{george-biddell-airy.jpg}
%\caption{George Biddell Airy FRS (1801 - 1892) }
\end{figure}
\end{center}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.6. }

\begin{itemize}\itemsep0.5em

\item  {\color{red} 例子7.2.3：求勒让德方程在原点附近的幂级数解：$$(1-x^2)y''-2xy'+n(n+1)y=0.$$ } 

\vspace{-0.5cm}

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  这里 $p(x)=\frac{-2x}{1-x^2}$ 和 $q(x)=\frac{n(n+1)}{1-x^2}$ 在 $-1<x<1$ 可以展开成收敛幂级数，所以勒让德方程在这个区间有幂级数解。

\item  设幂级数解为 $y=C_0 + C_1x + C_2x^2 + C_3x^3+\cdots$. 

\item  代入方程，得到递推关系式
$$(k+2)(k+1)C_{k+2} + (n+k+1)(n-k)C_k=0. $$

\item  得到两个线性无关的幂级数解：奇数次项幂级数和偶数次项幂级数。两个幂级数解的前四项为：


\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.2.7. Adrien Marie Legendre (1752 - 1833) }

\begin{center}
\begin{figure}
\includegraphics [height=0.65\textheight, width=0.3\textwidth]{adrien-marie-legendre.jpg}
%\caption{Adrien Marie Legendre (1752 - 1833) }
\end{figure}
\end{center}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.3.1. 勒让德多项式}

\begin{itemize}

\item  {\color{red} 问题：什么是勒让德多项式？}

\item  解答：当 $n$ 是非负整数时，勒让德方程 $$(1-x^2)y''-2xy'+n(n+1)y=0$$ 有多项式解
$$
P_n (x) = \frac{1}{2^n} \sum\limits_{k=0}^{\left[\frac{n}{2}\right]} \frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!} x^{n-2k}, 
$$
这些多项式称为勒让德多项式。


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.3.2. 勒让德多项式}

\begin{itemize}

\item  {\color{red} 问题：写出一些勒让德多项式。}

\item  解答：

\vspace{-0.5cm}

\begin{eqnarray*}
P_0(x) &=& 1 \\
P_1(x) &=& x \\
P_2(x) &=& \frac{1}{2}(3x^2-1) \\
P_3(x) &=& \frac{1}{2}(5x^3-3x) \\
P_4(x) &=& \frac{1}{8}(35x^4-30x^2+3) \\
P_5(x) &=& \frac{1}{8}(63x^5-70x^3+15x) 
%\cdots && \cdots
\end{eqnarray*}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.3.3. 勒让德多项式的性质1}

\begin{itemize}
\item  {\color{red} 问题：证明勒让德多项式满足罗德里格斯公式，
$$P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n} (x^2-1)^n.$$ 
}

\item  证明：先将 $(x^2-1)^n$ 用二项式公式展开，可得
$$ (x^2-1)^n = \sum\limits_{k=0}^{n} \frac{n!}{k!(n-k)!} (x^2)^{n-k}(-1)^k. 
$$
然后对每一项求导 $n$ 次，发现结果与勒让德多项式相差一个常数倍。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{7.3.4. 勒让德多项式的性质2}

\begin{itemize}
\item  {\color{red} 问题：证明勒让德函数系在区间 $[-1,1]$ 上是相互正交的，即 
\begin{eqnarray*}
\int_{-1}^{1} P_n(x)P_m(x)dx = 
\left\{ \begin{array}{ll}
0, & m\neq n, \\
\sigma_n>0, & m=n. 
\end{array}\right.
\end{eqnarray*}

}

\item  证明：使用罗德里格斯公式，然后多次使用分部积分法。

\end{itemize}

\end{frame}

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\begin{frame}{7.3.5. 勒让德多项式的性质2（续）}

\begin{itemize}
\item  {\color{red} 问题：什么是广义傅立叶级数？}

\item  解答：若函数 $f(x)$ 在区间 $[-1,1]$ 上是可积的，那么可以研究广义傅里叶级数 
$$f(x) \sim \sum\limits_{n=0}^{\infty} a_nP_n(x),$$ 
其中广义傅里叶系数定义为
$$a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x)P_n(x)dx.$$

\end{itemize}

\end{frame}

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\begin{frame}{7.3.6. 勒让德多项式的性质2（续）}

\begin{itemize}
\item  {\color{red} 问题：什么时候由勒让德函数系定义的广义傅立叶级数是收敛的？}

\item  解答：
\begin{enumerate}
\item  函数 $(1-x^2)^{-1/4}f(x)$ 在区间 $[-1,1]$ 是绝对可积的。
\item  狄利克雷条件、或迪尼条件、或赫尔德条件成立。
\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{7.4.1. 广义幂级数解法：例子7.4.1}

\begin{itemize}\itemsep0.5em

\item  {\color{red}问题：求微分方程 $x^2y''-2y=0$ 在奇点 $x=0$ 附近的解。}

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  猜想解函数具有形式 $y=x^\rho$, 代入方程可得 
$$
x^2(x^\rho)'' -2x^\rho = 0. 
$$

\item  求导，合并同类项，可得 $$[\rho(\rho-1)-2] x^\rho =0.$$

\item  求得 $\rho = 2$ 或 $\rho=-1$. 

\item  通解为 $y=C_1x^2+\frac{C_2}{x}$, 其中 $C_1,C_2$ 是任意常数。 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{7.4.2. 例子7.4.2 }

\begin{itemize}

\item  {\color{red}问题：讨论微分方程 $x^2y'' +(3x-1)y'+y=0$ 在奇点 $x=0$ 附近存在幂级数解的可能性。}

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  设有幂级数解 $y=C_0 + C_1x + C_2x^2 + \cdots + C_n x^n + \cdots$. 
\item  代入方程，可得递推公式 $$C_{n+1} = (n+1)C_n.$$
\item  得到形式幂级数解为 $$y = C_0\sum\limits_{n=0} ^{\infty} n! x^n. $$
\item  这个幂级数对任意 $x\neq 0$ 都是发散的。

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{7.4.3. 例子7.4.3 }

\begin{itemize}\itemsep0.5em

\item  {\color{red}问题：求微分方程 $x^2y''+xy'+(x^2-1/4)y=0$ 在奇点 $x=0$ 附近的解。}

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  先作变量代换 $y=\frac{u}{\sqrt{x}}$. 将 $y', y''$ 换成 $u', u''$:  
{\footnotesize 
\begin{eqnarray*}
\frac{dy}{dx} &=& \frac{\sqrt{x}\frac{du}{dx} - u\frac{1}{2\sqrt{x}}}{x} = \frac{2xu' - u}{2x\sqrt{x}}, \\ 
\frac{d^2y}{dx^2} &=&  
\end{eqnarray*}
}
\item  原方程化为 $u''+u=0$. 得到两个解函数 $u=\cos(x)$ 和 $u=\sin(x)$. 

\item  原方程的解函数为 $y=\frac{\cos(x)}{\sqrt{x}}$ 和 $y=\frac{\sin(x)}{\sqrt{x}}$. 

\item  这两个函数可以展开成广义幂级数的形式。

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{7.4.4. 正则奇点}

\begin{itemize}

\item  {\color{red}问题：什么是正则奇点？}

\item  解答：
设函数 $P(x), Q(x), R(x)$ 都是多项式，那么称 $x_0$ 是微分方程 $$(x-x_0)^2P(x)y'' + (x-x_0)Q(x)y'+R(x)y=0$$ 的正则奇点。


\end{itemize}

\end{frame}

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\begin{frame}{7.4.5. 正则奇点附近存在广义幂级数解}

\begin{itemize}

\item  {\color{red}定理7.3：微分方程在正则奇点附近有收敛的广义幂级数解：
$$y=\sum\limits_{k=0}^{\infty} C_k (x-x_0)^{k+\rho}.$$
其中 $C_0\neq 0$. 指标 $\rho$ 和系数 $C_k$ 用待定系数法求得。
}

\end{itemize}

\end{frame}

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\begin{frame}{7.4.6. 存在广义幂级数解的证明}

\begin{enumerate}

\item  设 $x_0=0$. 在 $|x|<r$ 时，将微分方程的系数都写成幂级数形式，
$$x^2y'' + x\left(\sum\limits_{k=0}^\infty a_kx^k\right)y' + \left(\sum\limits_{k=0}^\infty b_kx^k\right)y = 0. $$

\item  设广义幂级数解为 $y=\sum\limits_{k=0}^{\infty} C_k x^{k+\rho}$. 
代入然后比较系数，可得指标方程 $$\rho(\rho-1)+a_0\rho+b_0=0. $$ 
 
\item  计算指标方程的解，计算形式幂级数的每个系数。

\item  证明形式幂级数解在 $0<|x|<r_3<r$ 是收敛的。


\end{enumerate}

\end{frame}

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\begin{frame}{7.4.7. 例子7.4.4 }

\begin{itemize}

\item  {\color{red} 问题：求解贝塞尔方程 $x^2y'' + xy' + (x^2-n^2)y = 0$, 其中常数 $n\ge 0$. }

\item  解答：

\begin{enumerate}\itemsep0.5em

\item  奇点 $x=0$ 是正则的。

\item  设广义幂级数解为 $y=\sum\limits_{k=0}^{\infty} C_k (x-x_0)^{k+\rho}$. 求得指标 $\rho=\pm n$. 

\item  求得第一类贝塞尔函数 $$J_n(x) = \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{\Gamma(n+k+1)\Gamma(k+1)} \left(\frac{x}{2}\right)^{2k+n}.$$

\item  对 $n$ 进行讨论。

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{7.5.1. 贝塞尔函数的性质}

\begin{enumerate}
\item  当 $x\to\infty$ 时 $J_n(x)$ 的渐近式。从变量代换 $y=\frac{u}{\sqrt{x}}$ 可得。
\item  贝塞尔函数系 $J_n(\beta_1t),J_n(\beta_2t),\cdots J_n(\beta_nt),\cdots $ 在区间 $0\le t\le 1$ 上的正交性质。

\begin{center}
\includegraphics [height=0.5\textheight, width=0.7\textwidth]{ode-example-7-5-1.png}
\end{center}

\end{enumerate}


\end{frame}

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\begin{frame}[fragile=singleslide]{7.5.2. }

{\footnotesize\color{blue}
\begin{python}
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as spl 

x = np.linspace(0,20,101)
y0 = spl.jv(0,x); y1 = spl.jv(1,x); y2 = spl.jv(2,x)

fig=plt.figure()
ax=fig.add_subplot(111)
ax.plot(x, y0, 'r-', label='n=0')
ax.plot(x, y1, 'g-', label='n=1')
ax.plot(x, y2, 'b-', label='n=2')
ax.legend(loc='best')
fig.savefig('ode-example-7-5-1.png')
\end{python}
}

\end{frame}

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\begin{frame}{7.5.3. Friedrich Wilhelm Bessel (1784 - 1846) }

\begin{center}
\includegraphics [height=0.66\textheight, width=0.55\textwidth]{friedrich-wilhelm-bessel-stamp.jpg}
\end{center}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{习题7.1. }

\begin{enumerate}
\item[2.]  （选做）设函数 $p(x),q(x)$ 可以在区间 $(-a,a)$ 中展开成收敛幂级数，证明微分方程初值问题
$$y'' + p(x)y' + q(x)y = 0, \,\, y(0)=y_0, y'(0)=y_0'$$
在区间 $(-a,a)$ 中存在收敛幂级数解而且是唯一的。

\end{enumerate}

\end{frame}

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\begin{frame}{习题7.2. }

\begin{enumerate}\itemsep0.5em
\item  求出微分方程在 $x_0$ 附近的两个线性无关的幂级数解，
\begin{enumerate}\itemsep0.5em
\item  $y'' -xy' - y = 0, \,\, x_0=0$.
\item  $y'' - xy' - y = 0, \,\, x_0=1$. 
\item  $(1-x)y'' + y = 0, \,\, x_0=0$. 
\end{enumerate}

\item  求出微分方程初值问题的幂级数解的前四项，
\begin{enumerate}\itemsep0.5em
\item  $y'' + xy' + y = 0, \,\, y(0)=1, y'(0)=0$.
\item  $y'' +\sin(x) y' +\cos(x) y = 0, \,\, y(0)=0, y'(0)=1$.
\end{enumerate}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{习题7.3. }

\begin{enumerate}
\item   （选做）证明函数 $$G(x,t) = \frac{1}{\sqrt{1-2xt+t^2}}$$  
关于 $t$ 的幂级数展开为 
$$\sum\limits_{n=0}^\infty P_n(x)t^n, $$
其中 $P_n(x)$ 是勒让德多项式。

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{习题7.4. }

\begin{enumerate}\itemsep0.5em
\item  判别 $x=-1,0,1$ 是下述微分方程的常点、正则奇点、非正则奇点：
\begin{enumerate}\itemsep0.5em
\item  $xy'' + (1-x)y' + xy = 0 $.
\item  $(1-x^2)y'' -2xy' +n(n+1)y = 0$. 
\item  $2x^4(1-x^2)y'' +2xy' +3x^3y = 0$. 
\item  $x^2(1-x^2)y'' + 2x^{-1}y' + 4y = 0$. 
%\item  $y '' +\frac{x^2}{(1+x)^2} y ' + 3(1+x)^2y = 0$. 
\end{enumerate}

\item  用广义幂级数求解下列微分方程：
\begin{enumerate}\itemsep0.5em
\item  $2xy'' + y' + xy = 0 $.
\item  $x^2y'' + xy' + (x^2-1/9)y = 0$. 
\end{enumerate}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

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\begin{frame}{7.5. Bessel Biography - University of St Andrews, Scotland}

\begin{itemize}

\item[1.]  Wilhelm Bessel's father was a civil servant in Minden. Bessel attended the Gymnasium in Minden for four years but he did not appear to be very talented, finding Latin difficult. The fact that he later became proficient in Latin, teaching himself the language, probably suggests that the Gymnasium failed to inspire Bessel. In January 1799, at the age of 14, he left school to become an apprentice to the commercial firm of Kulenkamp in Bremen. The firm was involved in the import-export business. 


\end{itemize}

\end{frame}

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\begin{frame}{7.5. Bessel Biography}

\begin{itemize}

\item[2.]  At first Bessel received no salary from the firm but, as his accounting skills became appreciated by the firm, he received a small salary. Interest in the countries his firm dealt with led Bessel to spend his evenings studying geography, Spanish and English. His interests turned towards navigation and he considered the problem of finding the position of a ship at sea. This in turn led him to study astronomy and mathematics and he began to make observations to determine longitude. 

\end{itemize}

\end{frame}

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\begin{frame}{7.5. Bessel Biography}


\begin{itemize}

\item[3.]  In 1804 Bessel wrote a paper on Halley's comet, calculating the orbit using data from observations made by Harriot in 1607. He sent his results to Heinrich Olbers, the leading comet expert of his time, who recognised at once the quality of Bessel's work and Olbers gave Bessel the task of making further observations to carry his work further. The resulting paper, at the level required for a doctoral dissertation, was published on Olbers' recommendation. From that time on Bessel concentrated on astronomy, celestial mechanics and mathematics. 

\end{itemize}

\end{frame}

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\begin{frame}{7.5. Bessel Biography}

\begin{itemize}

\item[4.]  Olbers suggested to Bessel, who was still an apprentice to the import-export firm, that he should become a professional astronomer. In 1806 he accepted the post of assistant at the Lilienthal Observatory, a private observatory near Bremen. It was only after some considerable thought that Bessel left the affluence that was guaranteed in his commercial job choosing instead the near poverty of the Observatory post. However the Lilienthal Observatory gave him valuable experience observing planets, in particular Saturn, its rings and satellites. He also observed comets and continued his study of celestial mechanics. In 1807 he began to work on reducing James Bradley's observations (Bradley was English Astronomer Royal from 1742 to 1762) of the positions of 3222 stars made around 1750 at Greenwich. 

\end{itemize}

\end{frame}

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\begin{frame}{7.5. Bessel Biography}

\begin{itemize}

\item[5]  Bessel's brilliant work was quickly recognised and both Leipzig and Greifswald offered him posts. However he declined both. In 1809, at the age of 26, Bessel was appointed director of Frederick William III of Prussia's new Königsberg Observatory and professor of astronomy. It was not possible for Bessel to receive a professorship without first being granted the title of doctor. A doctorate was awarded by the University of Göttingen on the recommendation of Gauss, who had met Bessel in Bremen in 1807 and recognised his talents. 

\end{itemize}

\end{frame}

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\begin{frame}{7.5. Bessel Biography}

\begin{itemize}

\item[6.]  Although the Observatory at Königsberg was still under construction, Bessel took up his new post on 10 May 1810. He continued to work on Bradley's observations while work continued on the observatory from 1810 to 1813. Bessel's work had now become known internationally and he was honoured with the award of the Lalande Prize from the Institut de France for his tables of refraction based on Bradley's observations. Also during this period, in 1812, he was elected to the Berlin Academy. 

\end{itemize}

\end{frame}

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